Skills Review: Integration By Parts
If you integrate both sides of the product rule and rearrange, then you get the integration by parts
formula:
Z
Z
udv = uv − vdu.
The method involves choosing u and dv, computing du and v, and using the formula. See the last page
of this review for some fast ways to use this formula.
Before we get started, let me remind you of the following integrals (these only require substitution):
Rule
Example
Z
Z
1
1
αt
αt
e dt = e + C
e5t dt = e5t + C
α
5
Z
Z
1
1
cos(αt) dt = sin(αt) + C
cos(3t) dt = sin(3t) + C
α
3
Z
Z
1
1
1
t dt = −4 cos
t +C
sin(αt) dt = − cos(αt) + C
sin
α
4
4
Now we can begin. Here are a few classic examples of integration by parts, try them out and see if you
can get the given answer (answers are on the right). These should be easy exercises for you, come ask
in office hours if they are not and I can give you a refresher:
Z
1
1
1.
te2t dt.
= te2t − e2t + C.
2
4
Z
1
1
2.
t sin(3t) dt.
= − t cos(3t) + sin(3t) + C.
3
9
Z
3.
ln(t) dt.
= t ln(t) − t + C.
Z
4.
t2 e5t dt.
1
2
2 5t
= t2 e5t − te5t +
e + C.
5
25
125
For what we are doing in this class, we need to be able to do the following types of problems:
Z
1
1
•
teαt dt = teαt − 2 eαt + C.
α
α
Z
1
2
2
•
t2 eαt dt = t2 eαt − 2 teαt + 3 eαt + C.
α
α
α
Z
1
3
6
6
•
t3 eαt dt = t3 eαt − 2 t2 eαt + 3 teαt + 4 eαt + C.
α
α
α
α
Never Ending Integration by Parts (and how to end it)
It also is important in this class that we can integrate the following examples (These require integration
by parts twice and a ‘trick’). Try these:
Z
1
1
1.
et cos(3t) dt.
Answer: te2t − e2t + C.
2
4
Z
1
1
2.
e2t sin(4t) dt.
Answer: − t cos(3t) + sin(3t) + C.
3
9
The general result is (you can quote these):
Z
β
α
•
eαt sin(βt) dt = − 2
eαt cos(βt) + 2
eαt sin(βt) + C
2
α +β
α + β2
Z
β
α
eαt cos(βt) + 2
eαt sin(βt) + C
•
eαt cos(βt) dt = 2
2
2
α +β
α +β
Here is the general proof of one of these formula. Note that we use integration by parts twice, then get
all the integrals on one side by adding (that is the key to ending this seeming never ending integration
by
Z
Z parts):
α
1 αt
αt
eαt cos(βt) dt
by parts: u = eαt , dv = sin(βt)
e sin(βt) dt = − e cos(βt) +
β
β
Z
1 αt
α αt
α2
= − e cos(βt) + 2 e sin(βt) − 2 eαt sin(βt) dt
by parts: u = eαt , dv = cos(βt)
β
β
β
2 R
Adding αβ 2 eαt sin(βt) dt to both sides yields
Z
α2
α
1
1+ 2
eαt sin(βt) dt = − eαt cos(βt) + 2 eαt sin(βt)
β
β
β
Multiplying both sides by β 2 gives
Z
2
2
β +α
eαt sin(βt) dt = −βeαt cos(βt) + αeαt sin(βt)
And dividing by α2 + β 2 gives
Z
eαt sin(βt) dt = −
α
β
αt
e
cos(βt)
+
eαt sin(βt)
2
2
2
2
α +β
α +β
The other formula can be done in an nearly identical way.
Different Ways To Use Integration by Parts
Most students write u and dv in the margins or in a table, then compute du and v. That works well
and is all you need for this course. However, as you go into later courses (like Math 309), you might
want faster and more compact ways to do integration by parts. Here are a couple of methods to try.
This is just for your own interest.
1. Use the formula directly:
Instead of making a table in the margins you can do the integration and differentiation “in-line”
by directly using the formula. For example:
Z
Z
Z
4 3x
1 3x
4
4
1 3x
3x
= xe −
e
e d(4x) = xe3x − e3x + C
4xe dx = 4x d
3
3
3
3
9
R
R
In the evaluation above, we just replace u = 4x, dv = e3x in u dv = uv − v du. Notice in the
first step, we integrated to find v and replaced it in the formula. Then in the second step, we
used the formula replacing u and v. This is not necessarily better than the table method, but it
is something you can do in one line. Try it out and see if you like it.
2. Tabular Integration by Parts:
If you need to do multiple steps of integration,
then it might be helpful to make a big table of
R 4
derivatives and integrals. For example: x sin(x) dx will require many steps of integration by
parts. Rather than do each step separately, we can organize our work as follows:
u
dv
Sign
+
x4
sin(x)
−
4x3
cos(x)
2
+
12x − sin(x)
−
24x − cos(x)
+
24
sin(x)
0
cos(x)
Then we multiply u and v at each step with alternating signs (just like we would do if we actually
did each step with the formula). The final answer is:
Z
x5 sin(x) dx = x4 cos(x) − 4x3 (− sin(x)) + 12x2 (− cos(x)) − 24x sin(x) + 24 cos(x) + C